Question

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d₁ and dielectric constant k₁ and the other has thickness d₂ and dielectric constant k₂ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= d₁ + d₂) and effective dielectric constant k. The k is ______.

विकल्प

• `(k_1d_1 + k_2d_2)/(d_1 + d_2)`

• `(k_1d_1 + k_2d_2)/(k_1 + k_2)`

• `(k_1k_2 (d_1 + d_2))/((k_1d_1 + k_2d_2))`

• `(2k_1k_2)/(k_1 + k_2)`

Answer 1

A parallel plate capacitor is made of two dielectric blocks in series. One of the blocks has thickness d₁ and dielectric constant k₁ and the other has thickness d₂ and dielectric constant k₂ as shown in figure. This arrangement can be thought as a dielectric slab of thickness d (= d₁ + d₂) and effective dielectric constant k. The k is `underline((k_1k_2 (d_1 + d_2))/((k_1d_1 + k_2d_2)))`.

Explanation:

Here the system can be considered as two capacitors C₁ and C₂ connected in series as shown in figure.

The capacitance of parallel plate capacitor filled with dielectric block has thickness d₁ and dielectric constant K₂ is given by

`1/C_(eq) = 1/C_1 + 1/C_2` ⇒ `C_(eq) = (C_1C_2)/(C_1 + C_2)`

`C_(eq) = ((k_1ε_0A)/(d_1))/((k_1ε_0A)/(d_1)) ((k_2ε_0A)/(d_2))/((k_2ε_0A)/(d_2)) = (k_1k_2ε_0A)/(k_1d_2 + k_2d_1)`  ......(i)

We can write the equivalent capacitance as `C = (kε_0A)/(d_1 + d_2)`  ......(ii)

On comparing (i) and (ii) we have

`k = (k_1k_2(d_1 + d_2))/(k_1d_2 + k_2d_1)`